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reflection from the mirrors; a pair of rays signifying two rays which have originated at the same point of the source.

If the area of the luminous surface is sufficiently large the illumination at $P$ will be independent of the distance, form, or position of the surface. Suppose, therefore, that the luminous surface coincides with the surface $Om_{1}$. Its image in $Om_{1}$ will also coincide with $Om_{1}$, and its image in $Om_{2}$ will be a plane surface symmetrical with $Om_{1}$ with respect to the surface $Om_{2}$, and for every point, $p$, of the first image there is a corresponding point, $p'$, of the second, symmetrically placed and in the same phase of vibration. Suppressing, now, the source of light and the mirrors, and replacing them by the two images, the effect on any point, $P$, is unaltered.

Consider now a pair of points $pp'$. Let $\vartheta$ be the angle formed by the line joining $P$ and $p$ (or $p'$) with the normal to the surface; $\vartheta$ and $\varphi$ being both supposed small,

$\Delta =Pp'-Pp=pp'\cdot \cos \vartheta$.

The difference between this value of $\Delta$ and the true value is $2Pp\cdot \sin ^{2}{\tfrac {\psi }{2}}$, where $\psi$ is the angle subtended by $pp'$ at $P$. If $\vartheta$ is a small quantity, $\psi$ is a small quantity of the second order, and $\sin ^{2}{\tfrac {\psi }{2}}$ is a small quantity of the fourth order; consequently $2Pp\cdot \sin ^{2}{\tfrac {\psi }{2}}$ may be neglected. We have therefore, to a very close approximation, $\Delta =pp'\cos \vartheta$; or, substituting $2t$ for $pp'$, $2t$ being the distance between the images, at the point where they are cut by the line $Pp$,

$\Delta =2t\cos \vartheta$

Let $cdef$, $c'd'e'f'$, fig. 3, represent the two images, and let their intersection be parallel with $cf$, and their inclination be

Fig. 3

$2\varphi$. Let $P$ be the point considered; $P'$, the projection of $P$ on the surface $cdef$; and $PB$, the line forming with $PP'$ the angle